= Solution
For a finite-intensity <Poisson random measure>, <independence> of counts gives the exponential <characteristic function> formula. Apply it to the cutoff construction from part (a), including the deterministic compensation, and pass to the <L2 space> limit of the small jumps. The compensated integrand is $O(y^2)$ at zero and bounded on the large-jump region, so its integral converges absolutely:
$$
\mathbb E e^{iuX_t}=\exp\left\{t\int_{\mathbb R\setminus\{0\}}\left(e^{iuy}-1-iuy\mathbf1_{\{|y|\leq1\}}\right)c|y|^{-2}dy\right\}.
$$
Symmetry makes the absolutely integrable imaginary part zero. Substituting $z=|u|y$ in the real part and using the specified normalization gives
$$
2ct\int_0^\infty\frac{\cos(uy)-1}{y^2}dy
=-2ct|u|\int_0^\infty\frac{1-\cos z}{z^2}dz=-t|u|.
$$
Thus \b[$\mathbb E e^{iuX_1}=e^{-|u|}$], and this construction is the standard <Cauchy process>. To calculate the density, its integrable <characteristic function> permits <Fourier inversion>:
$$
f_{X_1}(x)=\frac1{2\pi}\int_{\mathbb R}e^{-|u|}e^{-iux}du
=\frac1\pi\operatorname{Re}\int_0^\infty e^{-(1+ix)u}du
=\boxed{\frac1{\pi(1+x^2)}}.
$$
This is the standard <Cauchy distribution>. The normalization also gives $c=1/\pi$: integration by parts turns $\int_0^\infty(1-\cos z)z^{-2}dz$ into $\int_0^\infty\sin z\,dz/z=\pi/2$. For completeness, with exponential damping the latter integral equals $\arctan(1/\eta)$, because its derivative in frequency $b$ is $\int_0^\infty e^{-\eta z}\cos(bz)dz=\eta/(\eta^2+b^2)$ and its value at $b=0$ is zero. Let $\eta\downarrow0$; integration by parts bounds the undamped and damped tails uniformly by a constant divided by the lower cutoff, justifying the limit.
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